terminal ray y Trigonometry Trigonometry is the study of triangles the relationship etween their sides and angles. Oddly enough our study of triangles egins with a irle. r 1 θ osθ P(x,y) s rθ sinθ x initial ray The unit irle is a irle with a radius equal to one. Notie that if we rotate upward from the x-axis an amount θ to a point P(x,y) on the irle and draw a line (terminal ray), we an onstrut a triangle. The legs of the triangle hange with rotation angle.
Trigonometry We define the radius (terminal ray) of the irle to e the hypotenuse of the triangle formed. As this hypotenuse rotates through the irle, the length of the sides opposite and adjaent to this angle hange in a presried way. What the Greeks did, was systematially θ measure this variation and reate three ratios to desrie it. These ratios they termed sine, osine, and tangent opposite opposite adjaent adjaent sin ( θ ) os ( θ ) ( ) radius hypotenuse radius hypotenuse tan θ y y opposite adjaent y θ sinθ x θ osθ x θ tanθ x
Trigonometry Why ratios? Why not just define the relationship etween the angle and the various sides of the triangle? Like sin(θ) vertial length, os(θ) horizontal length, and tan(θ) hypotenuse length Using ratios of the various sides allows the relationship to extend to any size irle (any length hypotenuse or leg). Trig funtions defined in terms of ratios give the same result for a given angle regardless of the size of the radius y y θ sinθ x θ sinθ x θ ( 45 ) 45 sin 0.707 1 θ ( 45 ) 45 sin 1.414
Trigonometry and Physis Great, ut how does knowing aout a relationship developed y some anient guys help me with physis? First, we need to understand that many physial quantities are what we all vetors. A vetor is any measurale quantity where diretion is important. Consider the following: Your friend alls and asks you to meet her two miles from the Westmoreland mall in half an hour. There are two measurale (physial) quantities that are important in this senario, the time (when) and the loation (where) you will meet. Sine time has only one diretion it need not e speified. Time is a salar quantity. Position on the other hand is a vetor and requires diretion to get it right.
Trigonometry and Physis mi mall Position or loation requires that you not only speify a magnitude ( mi.), ut a diretion from a referene point. There are a lot of plaes two miles from the mall, ut only one that is two miles, 15 degrees southwest of the mall.
Trigonometry and Physis N mall Notie that the displaement (hange in loation) is the radius (red) of a irle ( mi) E If I extend a vertial line (green) from the east-west axis down until it meets the radius, this gives me my southward motion and the vertial leg of a triangle mi If I now extend a line (lue) from the origin, along the eastwest axis, to the point on the east-west axis that the vertial line originated, I get the westward motion and the final leg of the triangle.
Trigonometry and Physis We know three things aout the triangle on the previous page and reprodued elow We know the length of the hypotenuse (displaement vetor) mi. We know the angle made etween the displaement vetor and the horizontal vetor 15 We know the angle etween the horizontal and vertial vetors 90 This is what is referred to as an angle-angle-side (AAS) onfiguration, and from it we an determine the lengths of the other two sides and the remaining angle. θ θ 15 mi
0.5 mi Trigonometry and Physis Starting with the unknown angle, you should all rememer from geometry that the sum of the interior angles of any triangle is 180. So if two of those angle, 15 and 90 add to 105, the final angle must e 180 105 or 75. Next, we an apply the trigonometri relationships (ratios) previously developed to solve for the remaining sides. Α 75 1.93 mi mi θ 15 Sine ( θ ) then opposite hypotenuse sin( θ ) While we solved for angle A first, we ould have started with any unknown quantity, the point is, if we know any three quantities we an solve for the remaining three. sin opposite hypotenuse ( ) mi sin(15 ) 0.5 mi opposite hypotenuse sin θ Sine os adjaent hypotenuse ( θ ) then adjaent hypotenuse os( θ ) ( ) mi os(15 ) 1.93 mi adjaent hypotenuse os θ
Useful Trigonometri Relationships Law of Sines. Law of Cosines. sin A a sin B a + sin C a os ( C ) Notie that if angle C is 90 (os(90 ) 0) the law of osines redues to the more familiar form of the Pythagorean equation. Pythagorean equation ONLY works with right triangles whereas the Law of Cosines works with any triangle + a A A B C C B a a
Let s Look at an Example Let s say you took a hike and set-off heading north, walking four kilometers. At that point you stopped for a drink and short rest. When you deided to ontinue, you veered right and walked three kilometers due east where you stopped for lunh. At the loation you took lunh, how far had you walked? How far where you from your starting point? N 4 km 3 km The total distane walked is the sum of the distane walked during eah interval, whih in this ase is 4 km and 3 km for a total of 7 km (4 km + 3km) The distane from the start alled the displaement, requires trigonometry
Let s Look at an Example N 3 km 4 km a B To find the displaement, we first reognize that the pattern of our motion forms the two legs of a right triangle with our displaement (green) eing the hypotenuse. From the Law of Cosines we an write: a + a os ( C ), ut C 90 so that, the length of our displaement redues to: a + a + 4 + 3 16 + 9 5 5 From the Law of Sines we determine sin( C ) sin( B) our aring, B: km
Let s Look at an Example N 4 km 3 km a B From the Law of Sines, we rearrange the equation solving for sin(b). ( C ) sin sin ( B) We need to undo the sine funtion, for that we use the arsine or sin -1. ( C ) 1 1 sin sin ( sin( B) ) sin NOTE sin -1 (x) is not the same as (1/sin(x)) The arsine undoes the sine so that sin -1 (sin(x)) is just x B sin 1 sin ( C ) 3sin( 90 ) sin 1 5 sin 1 3 5 36.87
Let s Look at an Example N 4 km 3 km a B As a hek, let s use the definition of the osine of an angle to evaluate our result for B. os ( θ ) adjaent adjaent hypotenuse or, hypotenuse os( θ ) adjaent ( 36.87 ) 4 km 5 km os This is indeed onsistent with the value of 4 km stated in the prolem.
Let s say you took a hike and set-off heading north, walking four kilometers. At that point you stopped for a drink and short rest. When you deided to ontinue, you veered right and walked three kilometers 15 north of east where you stopped for lunh. At the loation you took lunh, how far had you walked? How far where you from your starting point? N Let s Look at another Example 3 km 4 km a As efore, you ve walked a total distane of 7 km Your displaement however is not 5 km. It s more.
Looking at this other Example N 3 km 4 km a N θ 15 a E Oserve there are two right triangles one with a hypotenuse given y one with a hypotenuse given y
Looking at this other Example N a θ 15 os( 15 ) E sin( 15 ) a Notie that the ase of the green triangle,, is the same length as the ase of the red one. Notie that the height of the green triangle a is the sum of the length of vetor a and the height of the red triangle. a + a os 90 ( ) a a + sin 15 ( ) ( ) ( a + sin( 15 )) + ( os( 15 )) ( a + sin( 15 ))( os( 15 )) os( ) 90 ( ) os 15
Looking at this other Example ( a + sin( 15 )) + ( os( 15 )) ( a + sin( 15 ))( os( 15 )) os( ) 90 ( 4 + 3 sin( 15 )) + ( 3 os( 15 )) ( 4 + 3 sin( 15 ))( 3 os( 15 )) os( ) 90 0 sine os(90 ) 0 ( 4 + 3 sin( 15 )) + ( 3 os( 15 )) ( 4 + 0.78) + (.90) 5.59 km As expeted, your displaement is greater than 5 km
Graphial Look at Trig Funtions x in degrees NOTICE sin(90 ) 1 NOTICE os (90 ) 0